Answer
$\dfrac{1}{e}$
Work Step by Step
Here, we have $\ln f(x)=(\ln x) ^{-1/\ln x} $
and $\ln f(x)= \dfrac{-\ln x}{\ln x}$
or, $ f(x)=-1$
Hence, $e^{\lim\limits_{x \to 0^{+}} (-1)}=e^{-1}$
or, $e^{-1}=\dfrac{1}{e}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 55
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
